Question

An ice cute at 0.0*C was dropped into 30.0 g of water in a

Answer 1

Answer: The mass of ice cube is 77.90 grams

Explanation:

When ice is mixed with water, the amount of heat released by ice will be equal to the amount of heat absorbed by water.

$Heat_{\text{absorbed}}=Heat_{\text{released}}$

The equation used to calculate heat released or absorbed follows:

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

$m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]$       ......(1)

where,

$m_1$ = mass of ice = ? g

$m_2$ = mass of water = 30.0 g

$T_{final}$ = final temperature = 19.5°C

$T_1$ = initial temperature of ice = 0.0°C

$T_2$ = initial temperature of water = 45.0°C

$c_1$ = specific heat of ice = 2.108  J/g°C

$c_2$ = specific heat of water = 4.186 J/g°C

Putting values in equation 1, we get:

$m_1\times 2.108\times (19.5-0)=-[30.0\times 4.186\times (19.5-45.0)]$

$m_1=77.90g$

Hence, the mass of ice cube is 77.90 grams